The Stacks project

Lemma 43.22.2. Let $X \to P$ be a closed immersion of nonsingular varieties. Let $C' \subset P \times \mathbf{P}^1$ be a closed subvariety of dimension $r + 1$. Assume

  1. the fibre $C = C'_0$ has dimension $r$, i.e., $C' \to \mathbf{P}^1$ is dominant,

  2. $C'$ intersects $X \times \mathbf{P}^1$ properly,

  3. $[C]_ r$ intersects $X$ properly.

Then setting $\alpha = [C]_ r \cdot X$ viewed as cycle on $X$ and $\beta = C' \cdot X \times \mathbf{P}^1$ viewed as cycle on $X \times \mathbf{P}^1$, we have

\[ \alpha = \text{pr}_{X, *}(\beta \cdot X \times 0) \]

as cycles on $X$ where $\text{pr}_ X : X \times \mathbf{P}^1 \to X$ is the projection.

Proof. Let $\text{pr} : P \times \mathbf{P}^1 \to P$ be the projection. Since we are proving an equality of cycles it suffices to think of $\alpha $, resp. $\beta $ as a cycle on $P$, resp. $P \times \mathbf{P}^1$ and prove the result for pushing forward by $\text{pr}$. Because $\text{pr}^*X = X \times \mathbf{P}^1$ and $\text{pr}$ defines an isomorphism of $C'_0$ onto $C$ the projection formula (Lemma 43.22.1) gives

\[ \text{pr}_*([C'_0]_ r \cdot X \times \mathbf{P}^1) = [C]_ r \cdot X = \alpha \]

On the other hand, we have $[C'_0]_ r = C' \cdot P \times 0$ as cycles on $P \times \mathbf{P}^1$ by Lemma 43.17.1. Hence

\[ [C'_0]_ r \cdot X \times \mathbf{P}^1 = (C' \cdot P \times 0) \cdot X \times \mathbf{P}^1 = (C' \cdot X \times \mathbf{P}^1) \cdot P \times 0 \]

by associativity (Lemma 43.20.1) and commutativity of the intersection product. It remains to show that the intersection product of $C' \cdot X \times \mathbf{P}^1$ with $P \times 0$ on $P \times \mathbf{P}^1$ is equal (as a cycle) to the intersection product of $\beta $ with $X \times 0$ on $X \times \mathbf{P}^1$. Write $C' \cdot X \times \mathbf{P}^1 = \sum n_ k[E_ k]$ and hence $\beta = \sum n_ k[E_ k]$ for some subvarieties $E_ k \subset X \times \mathbf{P}^1 \subset P \times \mathbf{P}^1$. Then both intersections are equal to $\sum m_ k[E_{k, 0}]$ by Lemma 43.17.1 applied twice. This finishes the proof. $\square$

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